8.27 Calculate the work for each of the following processes beginning with a gas sample in a piston assembly with
T = 305 K, P = 1.79 atm, and V = 4.29 L: (a) irreversible expansion against a constant external pressure of 1.00 atm
to a final volume of 6.52 L; (b) isothermal, reversible expansion to a final volume of 6.52 L.
Someone already asked for clarification on this question, but I really do not understand how to do part b, as I cannot calculate the moles. I have n=(PV)/(RT) from the ideal gas law equation. I plugged in P=1.79 atm, V=6.52 L, R= 8.206x102 L.atm.K-1mol-1, and T=305 K. My answer is .00406. what am I doing wrong?
6th edition 8.27 part b
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Re: 6th edition 8.27 part b
The volume should be 4.29L, not the final volume, since you want to find the number of moles at the start of the expansion.
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Re: 6th edition 8.27 part b
Felicia1E wrote:The volume should be 4.29L, not the final volume, since you want to find the number of moles at the start of the expansion.
I plug this in and my answer is 3.068 x 10-5 mols. Plugging in the rest of the numbers, I get the work to be -0.3J which is very off from the -326 it is supposed to be. What am I doing incorrectly
Re: 6th edition 8.27 part b
Hmmm, well for the number of mols you get from PV=nRT using the values discussed should be .307 mols. I am not sure how you got 3.068 x 10^-5 mols. After getting the mols though, since it is a reversible isothermal expansion, you use w=-nRTln(Vf/Vi). If n=.307 mols, R=8.314 J/(mol K), T=305 K and Vf=6.52 and Vi=4.29, w=-326 J.
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